Circumference and area of ellipses
in non-Euclidean geometries
Jan Kärrman
Department of Mathematics
Uppsala University
Sweden
Abstract
In this paper we derive formulas for circumference and area of an ellipse
in hyperbolic geometry. Corresponding formulas for elliptic geometry are given
without proof. We also find a relation connecting the formulas for circumference
and area.
1
Preliminaries
This is a term paper for the course Geometry D given at Uppsala University. Due
to this nature of the document, derivation of the main formulas will be carried out
in quite some detail.
We are deriving the formulas under the simplified assumption that the curvature
constant k = 1.
All real numbers in√this paper will be non-negative; there will be several simplifications of the type x2 = x without further comments about the correctness of
such equations.
A well-known property of ellipses in Euclidean geometry is used to define a nonEuclidean ellipse:
Definition 1. Given two points P and Q, and a positive real number l > d = PQ.
The set of all points R such that PR + QR = l is called an ellipse with focal points
P and Q, and eccentricity d/l.
2
Ellipses in hyperbolic geometry
Given points P and Q, and a real number l, satisfying the conditions for an ellipse.
Let O be the midpoint of PQ. Let X be any point distinct from O, and let α = QOX.
−−→
Dedekind’s axiom gives that OX cuts the ellipse in exactly one point R=R(α). Let
r = r(α) = OR and s = QR, then PR = l−s, see Figure 1. Choose one of the sides of
←−→
the line PQ , and consider rays emanating from O on that side, with 0 α π/2.
1
R
l-s
X
r
s
α
P
d/2
O
Q
d/2
Figure 1
We have
d
d
− sinh r sinh cos α
2
2
d
d
cosh (l − s) = cosh r cosh − sinh r sinh cos (π − α)
2
2
d
d
= cosh r cosh + sinh r sinh cos α,
2
2
cosh s = cosh r cosh
which gives
d
2
d
cosh (l − s) − cosh s = 2 sinh r sinh cos α.
2
cosh (l − s) + cosh s = 2 cosh r cosh
Combining this with the trigonometrical identities
l − 2s
l
cosh
2
2
l − 2s
l
,
cosh (l − s) − cosh s = 2 sinh sinh
2
2
cosh (l − s) + cosh s = 2 cosh
we get
cosh
cosh r cosh d2
l − 2s
=
2
cosh 2l
sinh
sinh r sinh d2 cos α
l − 2s
=
,
2
sinh 2l
from which follows
1 = cosh2
=
l − 2s
l − 2s
− sinh2
2
2
sinh2 2l cosh2 r cosh2
⇒ sinh2
d
2
− cosh2 2l sinh2 r sinh2 d2 cos2 α
sinh2 2l cosh2
l
2
l
l
l
d
l
d
cosh2 = sinh2 (sinh2 r + 1) cosh2 − cosh2 sinh2 r sinh2 cos2 α
2
2
2
2
2
2
2
⇒ sinh2 r =
sinh2 2l (cosh2
sinh2 2l cosh2
d
2
− cosh2 d2 )
l
2
− cosh2 2l sinh2 d2 cos2 α
.
(1)
We define the major and minor axes of the ellipse in analogy with the Euclidean
case: set a = r(0), b = r(π/2), then
sinh2 a =
(cosh2 2l
sinh2 2l (cosh2
−
1) cosh2 d2
cosh2 b = sinh2 b + 1 =
l
2
−
− cosh2 d2 )
cosh2 2l (cosh2 d2
sinh2 2l (cosh2
l
2
− 1)
− cosh2 d2 )
sinh2 2l cosh2
d
2
= sinh2
+1=
l
2
cosh2
l
2
.
cosh2 d2
So we have
l = 2a,
cosh
cosh a
d
=
.
2
cosh b
It is easy to see that a and b can be use as parameters instead of l and d. Then
equation (1) becomes
2
a
sinh2 a cosh2 a − cosh
2
cosh b
,
sinh2 r =
2 cosh2 a
2
cosh2 a
sinh a cosh2 b − cosh a cosh2 b − 1 cos2 α
which, after multiplying numerator and denominator with the factor
becomes
sinh2 r =
cosh2 b − 1
1−
cosh2 a−cosh2 b
sinh2 a
cos2 α
=
sinh2 b
1−
sinh2 a−sinh2 b
sinh2 a
cos2 α
=
cosh2 b
,
sinh2 a cosh2 a
sinh2 b
(2)
1 − k12 cos2 α
where
sinh2 a − sinh2 b
.
k1 =
sinh a
Note that b r(α) a, and that r(α) is strictly decreasing for 0 α b < a.
Ri+1
π
2
when
δi
Ri
ri+1
ri
π/2n
O
Figure 2
We will first derive a formula for the circumference of an ellipse. Let n be a positive
πi
πi
integer, ri = r( 2n
), Ri = R( 2n
), and δi = Ri Ri+1 , see Figure 2. Then
π
πi
π 1
,
ri+1 = r
+
= ri +
ri + O
2n 2n
2n
n2
3
which gives
sinh ri+1 = sinh ri + O
1
,
n
and
π
cosh δi = cosh ri cosh ri+1 − sinh ri sinh ri+1 cos
2n
π
= cosh ri cosh ri+1 − sinh ri sinh ri+1 1 − 2 sin2
4n
2 π
+ cosh (ri+1 − ri )
= 2 sinh ri sinh ri+1 sin
4n
1
2
2 π
+ cosh (ri+1 − ri ) + O
= 2 sinh ri sin
4n
n3
π 2
π 1
1
2
ri + O
=2
sinh ri + cosh
+O
2
4n
2n
n
n3
π 2
1 π 2
1
.
=2
ri + O
sinh2 ri + 1 +
4n
2 2n
n3
From cosh δi = 2 sinh2 δ2i + 1, we get
π 2 1
2 δi
2
2
sinh ri + (ri ) + O
=
sinh
2
4n
n3
δi
π
1
⇒ sinh =
sinh2 ri + (ri )2 + O
2
4n
n2
π
1
2
2
⇒ δi =
.
sinh ri + (ri ) + O
2n
n2
Now sum all δi , i = 0, . . . , n − 1, and pass to the limit as n → ∞. This gives, by
symmetry, one fourth of the circumference C of the ellipse. Hence
π/2
sinh2 r(α) + r (α)2 dα.
C=4
0
From (2) we get
cosh r(α)r (α) =
− 3
d
sinh r(α) = −k12 sinh b sin α cos α 1 − k12 cos2 α 2 ,
dα
whence
sinh2 r(α) + r (α)2
k14 sinh2 b sin2 α cos2 α
sinh2 b
+
3
1 − k12 cos2 α
1 − k12 cos2 α cosh2 r(α)
k14 sinh2 b 1 − cos2 α cos2 α
sinh2 b
=
+
sinh2 b
1 − k12 cos2 α 1 − k 2 cos2 α3
+
1
2
2
1
1−k1 cos α
2
2
2
2
sinh b 1 − k1 cos α cosh b − k12 cos2 α + k14 sinh2 b 1 − cos2 α cos2 α
=
2 1 − k12 cos2 α
cosh2 b − k12 cos2 α
2
2
2
2
2
2 cosh b − k1 cosh b + 1 − k1 cos α
= sinh b 2 ,
1 − k12 cos2 α
cosh2 b − k12 cos2 α
=
4
and hence
π/2
C = 4 sinh b
0
cosh2 b − k12 cosh2 b + 1 − k12 cos2 α
dα.
2
2
2
2
2
cosh b − k1 cos α
1 − k1 cos α
This integral can be transformed into a more standardized form, by applying a
change of variable
α = arctan (u tan β),
where the real number u will be chosen later in some suitable way. Then
cos2 α =
1
1
cos2 β
1 − sin2 β
=
=
=
.
1 + tan2 α
1 + u2 tan2 β
cos2 β + u2 sin2 β
1 − (1 − u2 ) sin2 β
For convenience, we define
cβ =
1
,
1 − (1 − u2 ) sin2 β
m1 =
tanh2 a − tanh2 b
.
tanh2 a
We will also make use of the following equation
1
cosh2 a − cosh2 b
1
1
=
−
m1 =
tanh2 a cosh2 b cosh2 a
sinh2 a cosh2 b
=
k12
sinh2 a − sinh2 b
=
.
sinh2 a cosh2 b
cosh2 b
To avoid getting huge expressions; we apply the change of variable to the different
parts of the integral in turn.
i) cosh2 b − k12 cosh2 b + 1 − k12 cos2 α
1 − sin2 β
= cosh2 b − k12 cosh2 b + 1 − k12
1 − (1 − u2 ) sin2 β
= cβ cosh2 b 1 − (1 − u2 ) sin2 β − k12 (cosh2 b + 1 − k12 )(1 − sin2 β)
= cβ (1 − k12 )(cosh2 b − k12 ) + (k12 (1 − k12 ) − (1 − k12 − u2 ) cosh2 b) sin2 β)
Now choose u so that the term containing sin2 β vanishes
= cβ
sinh2 b
sinh2 b cosh2 b tanh2 b
sinh4 b
2
b(1
−
m
)
=
c
cosh
=
c
.
1
β
β
sinh2 a
sinh2 a tanh2 a
sinh2 a tanh2 a
That is, u has been chosen such that
(1 − k12 − u2 ) cosh2 b = k12 (1 − k12 )
(1 − k12 ) cosh2 b − k12 (1 − k12 )
⇒u=
= (1 − k12 )(1 − m1 )
cosh2 b
sinh b tanh b
sinh2 b tanh2 b
.
=
=
2
2
sinh a tanh a
sinh a tanh a
5
1 − sin2 β
2
2
2
2
2 = cβ 1 − k1 − (1 − k1 − u ) sin β
2
1 − (1 − u ) sin β
k12 (1 − k12 )
sinh2 b
2
2
2
sin
β
=
c
= cβ 1 − k1 −
β
2
2 (1 − m1 sin β).
cosh b
sinh a
ii) 1 − k12 cos2 α = 1 − k12
iii)
1 − sin2 β
1 − (1 − u2 ) sin2 β
= cβ cosh2 b − k12 − ((1 − u2 ) cosh2 b − k12 ) sin2 β
= cβ cosh2 b − k12 − (k12 (1 − k12 ) + k12 cosh2 b − k12 ) sin2 β
= cβ cosh2 b − k12 1 − k12 sin2 β = cβ cosh2 b(1 − m1 )(1 − k12 sin2 β)
cosh2 b − k12 cos2 α = cosh2 b − k12
tanh2 b
sinh2 b
2
2
(1
−
k
(1 − k12 sin2 β).
sin
β)
=
c
β
1
tanh2 a
tanh2 a
u(1 + tan2 β)
u(cos2 β + sin2β)
u
iv) dα =
dβ
=
dβ
=
dβ
2
2
1 + (u tan β)
cos2 β + u2 sin β
1 − (1 − u2 ) sin2 β
= cβ cosh2 b
= cβ
sinh b tanh b
dβ.
sinh a tanh a
Putting the pieces together, we get
4
π/2
b
cβ sinhsinh
c sinh b tanh b
2
a tanh2 a β
dβ
C = 4 sinh b
sinh2 b
sinh2 b
2
2 sin2 β) sinh a tanh a
c
(1
−
m
sin
β)
c
(1
−
k
1
β sinh2 a
β tanh2 a
1
0
which, after cancelling factors, finally gives
Proposition 1. The circumference C of an ellipse, having major and minor axis
with lengths 2a and 2b respectively, is given by
tanh b
C = 4 sinh b
tanh a
π/2
0
dα
,
(1 − m1 sin2 α) 1 − k12 sin2 α
where
tanh2 a − tanh2 b
,
m1 =
tanh2 a
sinh2 a − sinh2 b
.
k1 =
sinh a
Of course, the integral is a complete elliptic integral of the third kind; using a
common notation, we can write:
C = 4 sinh b
tanh b (k1 , m1 ).
tanh a
Next we derive a formula for the area of an ellipse. Choose an even n, large
π
is smaller than the angle of parallelism at O, with respect to the
enough so that 2n
−−→
−−→
perpendicular to OR0 through R0 . Let mi be the perpendicular to ORi through Ri .
6
−−→
π
Then, since ri r0 = a, the two rays forming angles 2n
with ORi cut mi , making up
an isosceles triangle, see Figure 3. Let si and 2bi be the lengths of its sides and base
respectively. By symmetry, one fourth of the area of the ellipse is given by summing
the areas of all such triangles, i = 1, 3, 5 . . . , n − 1, and then letting n → ∞.
bi R
i
bi
ri
si
si
mi
π/2n
O
Figure 3
Let Ai be the area of each of the two right triangles that make up the isosceles
triangle, then
tanh
Ai
ri
bi
= tanh tanh
2
2
2
cosh si = cosh bi cosh ri
π
sinh si
sinh bi = sin
2n
π
π sinh2 si = sin2
cosh2 si − 1
2n
2n
π
cosh2 bi cosh2 ri − 1
= sin2
2n
2 π
sinh2 bi cosh2 ri + sinh2 ri
= sin
2n
π
π 2
sin2 2n
sinh2 ri
1
2
=
sinh
r
+
O
sinh2 bi =
i
2
2 π
4
2n
n
1 − sin 2n cosh ri
π
1
sinh ri + O
sinh bi =
2n
n3
1
cosh bi = 1 + O
n2
sinh bi
π
bi
1
=
sinh ri + O
tanh =
2
cosh bi + 1
4n
n3
π
π
ri
Ai
1
1
=
=
tanh sinh ri + O
(cosh ri − 1) + O
tanh
2
4n
2
n3
4n
n3
π
1
.
(cosh ri − 1) + O
Ai =
2n
n3
⇒ sinh2 bi = sin2
⇒
⇒
⇒
⇒
⇒
⇒
7
Summing and passing to the limit as n → ∞ gives the area A:
π/2
A=4
cosh r(α) dα − 2π,
0
which, by (2), is
π/2
A=4
0
2
sinh b
+ 1 dα − 2π = 4
1 − k12 cos2 α
π/2
0
cosh2 b − k12 cos2 α
dα − 2π.
1 − k12 cos2 α
Now we make a change of variable
α = arctan (v cot β).
Then
cos2 α =
1
sin2 β
1
sin2 β
=
=
=
1 + tan2 α
1 + v 2 cot2 β
sin2 β + v 2 cos2 β
v 2 + (1 − v 2 ) sin2 β
and
dα =
v
v(sin2β + cos2 β)
−v(1 + cot2 β)
dβ = − 2
dβ
=
−
dβ
1 + (v cot β)2
sin2β + v 2 cos2 β
v + (1 − v 2 ) sin2 β
which gives
0
A = −4v
π/2
sin2 β
cosh2 b − k12 v2 +(1−v
2 ) sin2 β
dβ − 2π
sin2 β
2
2
2
1 − k12 v2 +(1−v
2 ) sin2 β (v + (1 − v ) sin β)
cosh2 b(v 2 + (1 − v 2 ) sin2 β) − k12 sin2 β
dβ − 2π
= 4v
2 + (1 − v 2 ) sin2 β) v 2 + (1 − v 2 ) sin2 β − k 2 sin2 β
(v
1
0
π/2
v 2 cosh2 b + ((1 − v 2 ) cosh2 b − k12 ) sin2 β
dβ − 2π
= 4v
2 + (1 − v 2 ) sin2 β) v 2 − (k 2 + v 2 − 1) sin2 β
(v
1
0
π/2
cosh2 b + ((1 − v 2 ) cosh2 b − k12 )/v 2 sin2 β
4
dβ − 2π.
=
v
2 − 1) sin2 β) 1 − (1 − (1 − k 2 )/v 2 ) sin2 β
(1
+
(1/v
1
0
π/2
As before, the constant v is chosen so that sin2 β vanishes from the numerator,
that is
√
coth a
k12
tanh b
=
.
=
1 − m1 =
v = 1−
2
tanh
a
coth b
cosh b
Then
1 − k12
cosh2 b
sinh2 b tanh2 a
=
,
=
v2
sinh2 a tanh2 b
cosh2 a
and we can formulate
8
Proposition 2. The area A of an ellipse, having major and minor axis with lengths
2a and 2b respectively, is given by
coth b
A = 4 cosh b
coth a
π/2
0
dα
− 2π,
1 − m2 sin2 α
1 − k22 sin2 α
where
coth2 a − coth2 b
m2 =
,
coth2 a
cosh2 a − cosh2 b
.
k2 =
cosh a
The formula has been written in this particular way to emphasize the similarity
with the formula for circumference. From this similarity one can easily derive the
following relation:
πi
C(a, b) = i A πi
(3)
2 − a, 2 − b + 2π .
See next section for some comments about this equation, and the corresponding
formula for elliptic geometry.
As always, it is good practice to check formulas for special cases and limits, especially since these formulas differ significantly from their counterparts in Euclidean
geometry. In particular, we would expect the following relations to hold:
i) C(a, a) = 2π sinh a
ii) A(a, a) = 4π sinh2
a
2
iii) C(a, b) → 4a, as b → 0
iv) A(a, b) → 0, as b → 0
π/2 1 − (1 − t2 ) sin2 α dα, 0 < t < 1, a small
v) C(a, ta) ≈ 4a 0
vi) A(a, ta) ≈ πta2 , 0 < t < 1, a small
The first two conditions are equalities with formulas for circles in hyperbolic geometry (see [1], pp. 407–410). The last two show that, for infinitesimal ellipses, the
hyperbolic formulas reduce to their Euclidean counterparts.
Proof. We will not prove all these conditions in detail, for some only an indication
of a proof is given.
i)
π/2
C(a, a) = 4 sinh a
1 dα = 2π sinh a
0
ii)
π/2
a
1 dα − 2π = 2π(cosh a − 1) = 4π sinh2
A(a, a) = 4 cosh a
2
0
9
iii)
a
lim C(a, b) = substitute α = arctan sinh
cot β
sinh
b
b→0
π/2
1 − k12 sin2 β
dβ
= lim 4 tanh a cosh b
b→0
1 − k12 tanh2 a sin2 β
0
π/2
= 4 tanh a
0
cos β
dβ = {tanh a sin β = x}
1 − tanh2 a sin2 β
tanh
a
dx
= 4a.
1 − x2
=4
0
iv)
tanh b
lim A(a, b) = α = arctan tanh
a cot β
b→0
π/2
= lim 4 cosh b
b→0
0
tanh2 b + tanh2 a − tanh2 b sin2 β
dβ − 2π
1 − k22 tanh2 b + k22 sin2 β
π/2
1 dβ − 2π = 0.
=4
0
v)
C(a, ta) ≈ 4t2 a
π/2
0
dα
dα
(1 − (1 − t2 ) sin2 α)3/2
π/2
cot β
1 − (1 − t2 ) sin2 β dβ.
= α = arctan
= 4a
t
0
vi) Using series expansion one gets
4
A(a, ta) =
t
π/2
2
1 + 1−t
a2 sin2 α
5t2 − 2 2
2
1+
a
dα − 2π + O(a4 )
1−t2
2a2
2
6
1+
1−
sin α
t2
0
3
5t2 − 2 2 π
5t2 − 3t − 2 2
4
1+
a
t 1−
a − 2π + O(a4 )
=
t
6
2
6
= πta2 + O(a4 ).
10
3
Ellipses in elliptic geometry
The corresponding formulas for elliptic geometry are derived with very similar calculations. Only the resulting equations are given here:
tan b
C(a, b) = 4 sin b
tan a
π/2
0
1−
cot b
A(a, b) = 2π − 4 cos b
cot a
tan2
a−tan2
tan2 a
dα
b
sin2 α
1−
π/2
0
1−
cot2
C(a, b) + A π2 − a, π2 − b = 2π
a−cot2
cot2 a
sin2 a−sin2 b
sin2 a
dα
b
sin2 α
1−
sin2 α
cos2 a−cos2 b
cos2 a
sin2 α
(4)
The formulas were derived under the assumption b a. But with a change of
variable α = π2 − β one can show that C(a, b) = C(b, a) and A(a, b) = A(b, a). So
equation (4), the counterpart of (3) in hyperbolic geometry, gives a relation between
circumference and area of two related ellipses; the two terms add up to the area of
the elliptic room. On a sphere with radius R the equation becomes
2
πR
C(a, b) R + A πR
2 − a, 2 − b = 2πR ,
which equals (3) for R = i, illustrating the fact that a sphere with imaginary radius
can be used as a model for hyperbolic geometry (see [1], p. 186).
Note that the formula for area should be written as A(b, a) to make its integral
an elliptic of the third kind. The form chosen was for the sake of symmetry with the
hyperbolic case.
It is not difficult to show that, in the standard unit sphere model of elliptic geometry, the orthogonal projection of an ellipse onto the plane tangent to the middle
point O, is a Euclidean ellipse – and conversely for Euclidean ellipses with major
axis 1. Or equivalently: ellipses are parallel projections of circles in three-space,
having radius 1 and centre coinciding with the centre of the sphere, onto the
sphere. This in turn shows that, for the extremal case when a = π/2, the ellipse
consists of two lines.
Formulas similar to (4) can be found for other objects in elliptic geometry. Let
C(h) and A(h) be the circumference and area respectively of an equilateral triangle
with altitude h. Then one can show that C(h)+A(π−h) = 2π. It would be interesting
to investigate such equations further, but that is beyond the scope of this document.
References
[1] Marvin J. Greenberg. Euclidean and non-Euclidean geometries. W. H. Freeman
and Company, New York, 1999.
11